Strong convergence theorems for fixed point problems of infinite family of asymptotically quasi-ϕ-nonexpansive mappings and a system of equilibrium problems

Abstract

In this paper, we introduce a general iterative algorithm for finding a common element of the set of common fixed points of infinite family of asymptotically quasi-ϕ-nonexpansive mappings and of the set of solutions for finite equilibrium problems in a real Banach space. Our results are the generalization of the results (Shehu in Comput. Math. Appl. 63:1089-1103, 2012; Kim in Fixed Point Theory Appl., 2011, doi:10.1186/1687-1812-2011-10) and (Kim and Buong in Fixed Point Theory Appl., 2011, doi:10.1155/2011/780764), and improvement of the result (Yang et al. in Appl. Math. Comput. 218:6072-6082, 2012).

MSC:47H09, 47H10, 47H17.

1 Introduction

Let C be a nonempty, closed and convex subset of a real Banach space E. A mapping $T:C\to C$ is called to be nonexpansive if

$$\parallel Tx-Ty\parallel \le \parallel x-y\parallel ,\mathrm{\forall }x,y\in C.$$

(1.1)

A mapping $T:C\to C$ is called to be quasi-nonexpansive if

$$\parallel Tx-x^{\ast }\parallel \le \parallel x-x^{\ast }\parallel ,\mathrm{\forall }x\in C,x^{\ast }\in F(T).$$

Let F be a bifunction of $C\times C$ into ℝ. The equilibrium problem is to find $x\in C$ such that

$$F(x,y)\ge 0,\mathrm{\forall }y\in C.$$

(1.2)

The set of solutions to equilibrium problem (1.2) is denoted by $EP(F)$. That is,

$$EP(F):=\{x\in C:F(x,y)\ge 0,\mathrm{\forall }y\in C\}.$$

Recently, Yang et al. [1] proved strong convergence theorems for approximation of common fixed points of countably infinite family of asymptotically quasi-ϕ-nonexpansive mappings in a uniformly smooth and strictly convex real Banach space, which has the Kadec-Klee property. More precisely, they proved the following theorem.

Theorem 1.1 Let E be a uniformly smooth and strictly convex Banach space, which has the Kadec-Klee property, and let C be a nonempty closed convex subset of E. Let G be a bifunction from $C\times C$ to ℝ satisfying (A1)-(A4), and let $T_i:C\to C$, $\mathrm{\forall }i\in \mathbb{N}$ be an infinite family of closed and asymptotically quasi-ϕ-nonexpansive mapping with $\{k_{ni}\}\subset [1,\mathrm{\infty })$, $k_{ni}\to 1$ as $n\to \mathrm{\infty }$, where $T_0=I$. Assume that $T_i$, $\mathrm{\forall }i\in \mathbb{N}$ is asymptotically regular on C and $\mathrm{\Im }={\bigcap }_{i=0}^{\mathrm{\infty }}F(T_i)\cap EP(G)$ is nonempty and bounded. Let $\{x_n\}$ be a sequence, generated by

$$\{\begin{array}{c}{x}_0\in E\mathit{\text{chosen arbitrarily,}}\hfill \\ C_1=C,\hfill \\ x_1={\mathrm{\Pi }}_{C_1}{x}_0,\hfill \\ y_n=J^{-1}\{{\sum }_{i=0}^{\mathrm{\infty }}{\alpha }_{ni}J{T}_i^n{x}_n\},\hfill \\ u_n\in C\mathit{\text{ such that }}G(u_n,y)+\frac{1}{r_n}〈y-u_n,J{u}_n-J{y}_n〉\ge 0,\mathrm{\forall }y\in C,\hfill \\ C_{n+1}=\{z\in C_n:\varphi (z,u_n)\le \varphi (z,x_n)+(k_n-1)M_n\},\hfill \\ x_{n+1}={\mathrm{\Pi }}_{C_{n+1}}{x}_0,\hfill \end{array}$$

where J is the duality mapping on E, $M_n=sup\{\varphi (z,x_n):z\in \mathrm{\Im }\}$ for each $n\ge 1$, $k_n={sup}_{i\ge 0}\{k_{ni}\}$, $\{r_n\}$ is real sequence in $[a,\mathrm{\infty })$, where a is some positive real number, $\{{\alpha }_{ni}\}$ is a real sequence in $[0,1]$ satisfying the following conditions: (a) ${\sum }_{i=0}^{\mathrm{\infty }}{\alpha }_{ni}=1$, $\mathrm{\forall }n\ge 1$, (b) ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\alpha }_{n0}{\alpha }_{ni}>0$, $\mathrm{\forall }i\in \mathbb{N}$. Then the sequence $\{x_n\}$ converges strongly to ${\mathrm{\Pi }}_{\mathrm{\Im }}{x}_0$.

In [2], Shehu introduced the following hybrid iterative scheme for approximating a common element of the set of fixed points of relatively quasi-nonexpansive mappings and the set of solutions to an equilibrium problem in a uniformly smooth and uniformly convex real Banach space: $x_0\in C$, $C_1=C$, $x_1={\mathrm{\Pi }}_{C_1}^f{x}_0$,

$$\{\begin{array}{c}{y}_n=J^{-1}({\alpha }_{n}J{x}_n+(1-{\alpha }_n)J{T}_n{x}_n),n\ge 1,\hfill \\ u_n=T_{r_{m,n}}^{F_m}{T}_{r_{m-1,n}}^{F_{m-1}}\cdots T_{r_{2,n}}^{F_2}{T}_{r_{1,n}}^{F_1}{y}_n,\hfill \\ C_{n+1}=\{w\in C_n:G(w,J{u}_n)\le G(w,J{x}_n)\},n\ge 1,\hfill \\ x_{n+1}={\mathrm{\Pi }}_{C_{n+1}}^f{x}_0,n\ge 1.\hfill \end{array}$$

Motivated by the facts above, the purpose of this paper is to prove a strong convergence theorem for finding a common element of the set of fixed points of asymptotically quasi-ϕ-nonexpansive mappings and the set of solutions to a system of equilibrium problems in a uniformly smooth and uniformly convex real Banach space, which has the Kadec-Klee property.

2 Preliminaries

Let E be a real Banach space, and let $E^{\ast }$ be the dual space of E. The duality mapping $J:E\to 2^{E^{\ast }}$ is defined by

$$J(x)=\{f\in E^{\ast }:〈x,f〉={\parallel x\parallel }^2={\parallel f\parallel }^2\}.$$

(2.1)

By Hahn-Banach theorem, $J(x)$ is nonempty.

The modulus of smoothness of E is the function ${\rho }_E:[0,\mathrm{\infty })\to [0,\mathrm{\infty })$ defined by

$${\rho }_E(\tau ):=sup\{\frac{1}{2}(\parallel x+y\parallel +\parallel x-y\parallel )-1:\parallel x\parallel \le 1,\parallel y\parallel \le \tau \}.$$

(2.2)

E is said to be uniformly smooth if ${lim}_{\tau \to 0}\frac{{\rho }_E(\tau )}{\tau }=0$.

Let $dimE\ge 2$. The modulus of convexity of E is the function ${\delta }_E:(0,2]\to [0,1]$ defined by

$${\delta }_E(ϵ):=inf\{1-\parallel \frac{x-y}{2}\parallel :\parallel x\parallel =\parallel y\parallel =1;ϵ=\parallel x-y\parallel \}.$$

(2.3)

E is said to be uniformly convex if $\mathrm{\forall }ϵ\in (0,2]$, there exists a $\delta =\delta (ϵ)>0$ such that for $x,y\in E$ with $\parallel x\parallel \le 1$, $\parallel y\parallel \le 1$ and $\parallel x-y\parallel \ge ϵ$, then $\parallel \frac{x+y}{2}\parallel \le 1-\delta$. Equivalently, E is uniformly convex if and only if ${\delta }_E(ϵ)>0$, $\mathrm{\forall }ϵ\in (0,2]$. E is strictly convex if for all $x,y\in E$, $x\ne y$, $\parallel x\parallel =\parallel y\parallel =1$, we have $\parallel \lambda x+(1-\lambda )y\parallel <1$, $\mathrm{\forall }\lambda \in (0,1)$.

It is well known that if E is uniformly smooth, then J is norm-to-norm uniformly continuous on each bounded subset of E. If E is smooth, then J is single-valued.

Recall that a Banach space E has the Kadec-Klee property if for any sequence $\{x_n\}\subset E$ and $x\in E$ with $x_n\rightharpoonup x$ and $\parallel x_n\parallel \to \parallel x\parallel$, then $\parallel x_n-x\parallel \to 0$, as $n\to \mathrm{\infty }$. It is well known that if E is a uniformly convex Banach space, then E has the Kadec-Klee property.

We denoted by ϕ the Lyapunov function from $E\times E$ to ℝ defined by

$$\varphi (x,y):={\parallel x\parallel }^2-2〈x,Jy〉+{\parallel y\parallel }^2,\mathrm{\forall }x,y\in E.$$

(2.4)

It follows from the definition of ϕ that

$${(\parallel x\parallel -\parallel y\parallel )}^2\le \varphi (x,y)\le {(\parallel x\parallel +\parallel y\parallel )}^2,\mathrm{\forall }x,y\in E.$$

(2.5)

Let E be a reflexive strictly convex and smooth Banach space. Then for $x,y\in E$, $\varphi (x,y)=0$ if and only if $x=y$ (see [3, 4]).

Definition 2.1 Let C be a nonempty closed convex subset of E, and let T be a mapping from C into itself. A point $p\in C$ is said to be an asymptotic fixed point of T if C contains a sequence $\{x_n\}$, which converges weakly to p and ${lim}_{n\to \mathrm{\infty }}\parallel x_n-T{x}_n\parallel =0$. The set of asymptotic fixed points of T is denoted by $\tilde{F}(T)$.

We say that T is a relatively nonexpansive mapping [5–8] if the following conditions are satisfied:

• (R1) $F(T)\ne \mathrm{\varnothing }$;

• (R2) $\varphi (p,Tx)\le \varphi (p,x)$, $\mathrm{\forall }x\in C$, $p\in F(T)$;

• (R3) $F(T)=\tilde{F}(T)$.

If T satisfies (R1) and (R2), then T is said to be relatively quasi-nonexpansive [9–11].

Definition 2.2 We say that T is an asymptotically ϕ-nonexpansive mapping if there exists a sequence $\{k_n\}\subset [1,\mathrm{\infty })$ with $k_n\to 1$ as $n\to \mathrm{\infty }$ such that $\varphi (T^{n}x,T^{n}y)\le k_n\varphi (x,y)$, $\mathrm{\forall }x,y\in C$. We say that T is an asymptotically quasi-ϕ-nonexpansive [11, 12] mapping if $F(T)\ne \mathrm{\varnothing }$ and there exists a sequence $\{k_n\}\subset [1,\mathrm{\infty })$ as $n\to \mathrm{\infty }$ such that $\varphi (p,T^{n}x)\le k_n\varphi (p,x)$, $\mathrm{\forall }x\in C$, $p\in F(T)$.

It is easy to see that the class of relatively quasi-nonexpansive mappings and asymptotically quasi-ϕ-nonexpansive mappings contains the class of relatively nonexpansive mappings. The class of asymptotically quasi-ϕ-nonexpansive mappings is more general than the class of relatively quasi-nonexpansive mappings.

Following Alber [13], the generalized projection ${\mathrm{\Pi }}_C:E\to C$ is defined by

$${\mathrm{\Pi }}_C(x)=\{u\in C:\varphi (u,x)=\underset{y\in C}{min}\varphi (y,x)\},\mathrm{\forall }x\in E.$$

The existence and uniqueness of the operator ${\mathrm{\Pi }}_C$ follows from the properties of the function $\varphi (y,x)$ and strict monotonicity of mapping J (see, for example, [3, 4, 13, 14]). If E is a Hilbert space, then $\varphi (y,x)={\parallel y-x\parallel }^2$, $x,y\in E$ and ${\mathrm{\Pi }}_C$ is the metric projection $P_C$ of E onto C.

Next, we recall the concept and properties of generalized f-projector operator. Let $G:C\times E^{\ast }\to \mathbb{R}\cup \{+\mathrm{\infty }\}$ be a function defined as follows:

$$G(\xi ,\phi )={\parallel \xi \parallel }^2-2〈\xi ,\phi 〉+{\parallel \phi \parallel }^2+2\rho f(\xi ),$$

where $\xi \in C$, $\phi \in E^{\ast }$, ρ is a positive number, and $f:C\to \mathbb{R}\cup \{+\mathrm{\infty }\}$ is proper, convex and lower semi-continuous. From the definitions of G and f, it is easy to see that the following properties hold:

• (i) $G(\xi ,\phi )$ is convex and continuous with respect to φ when ξ is fixed;

• (ii) $G(\xi ,\phi )$ is convex and lower semi-continuous with respect to ξ when φ is fixed.

Definition 2.3 [15]

Let E be a real Banach space with its dual $E^{\ast }$. Let C be a nonempty closed convex subset of E. We say that ${\mathrm{\Pi }}_C^f:E^{\ast }\to 2^C$ is a generalized f-projection operator if

$${\mathrm{\Pi }}_C^f\phi =\{u\in C:G(u,\phi )=\underset{\xi \in C}{inf}G(\xi ,\phi )\},\mathrm{\forall }\phi \in E^{\ast }.$$

Lemma 2.4 [15]

Let E be a reflexive Banach space with its dual $E^{\ast }$. Let C be a nonempty closed convex subset of E. Then the following statements hold:

• (i) ${\mathrm{\Pi }}_C^f\phi$ is a nonempty closed convex subset of C for all $\phi \in E^{\ast }$;

• (ii) If E is smooth, then for all $\phi \in E^{\ast }$, $x\in {\mathrm{\Pi }}_C^f\phi$ if and only if

  $$〈x-y,\phi -Jx〉+\rho f(y)-\rho f(x)\ge 0,\mathrm{\forall }y\in C;$$

• (iii) [16]If E is strictly convex, then ${\mathrm{\Pi }}_C^f$ is a single-valued mapping.

Recall that J is a single-valued mapping when E is a smooth Banach space. There exists a unique element $\phi \in E^{\ast }$ such that $\phi =Jx$ for each $x\in E$. This substitution in (2.3) gives

$$G(\xi ,Jx)={\parallel \xi \parallel }^2-2〈\xi ,Jx〉+{\parallel x\parallel }^2+2\rho f(\xi ).$$

Now, we consider the second generalized f-projection operator in Banach space.

Definition 2.5 Let E be a real Banach space and C be a nonempty closed convex subset of E. We say that ${\mathrm{\Pi }}_C^f:E\to 2^C$ is a generalized f-projection operator if

$${\mathrm{\Pi }}_C^{f}x=\{u\in C:G(u,Jx)=\underset{\xi \in C}{inf}G(\xi ,Jx)\},\mathrm{\forall }x\in E.$$

Obviously, the definition of relatively quasi-nonexpansive mapping T is equivalent to

• (R′1) $F(T)\ne \mathrm{\varnothing }$;

• (R′2) $G(p,JTx)\le G(p,Jx)$, $\mathrm{\forall }x\in C$, $p\in F(T)$.

Lemma 2.6 [17]

Let C be a nonempty, closed and convex subset of a smooth and reflexive Banach space E. Then the following statements hold:

• (i) ${\mathrm{\Pi }}_C^{f}x$ is a nonempty closed convex subset of C for all $x\in E$;

• (ii) For all $x\in E$, $\hat{x}\in {\mathrm{\Pi }}_C^{f}x$ if and only if

  $$〈\hat{x}-y,Jx-J\hat{x}〉+\rho f(y)-\rho f(\hat{x})\ge 0,\mathrm{\forall }y\in C;$$

• (iii) [16]If E is strictly convex, then ${\mathrm{\Pi }}_C^{f}x$ is a single-valued mapping.

Lemma 2.7 [18]

Let E be a Banach space, and $f:E\to \mathbb{R}\cup \{+\mathrm{\infty }\}$ is convex and lower semi-continuous. Then there exists $x^{\ast }\in E^{\ast }$ and $\alpha \in \mathbb{R}$ such that

$$f(x)\ge 〈x,x^{\ast }〉+\alpha ,\mathrm{\forall }x\in E.$$

Lemma 2.8 [17]

Let C be a nonempty closed convex subset of a smooth and reflexive Banach space E. Let $x\in E$ and $\hat{x}\in {\mathrm{\Pi }}_C^{f}x$. Then

$$\varphi (y,\hat{x})+G(\hat{x},Jx)\le G(y,Jx),\mathrm{\forall }y\in C.$$

Lemma 2.9 [1, 19]

Let E be a uniformly smooth and strictly convex Banach space, which has the Kadec-Klee property, and let C be a nonempty closed convex subset of E. Let T be a closed and asymptotically quasi-ϕ-nonexpansive mapping. Then $F(T)$ is a closed and convex subset of C.

Lemma 2.10 [1]

Let E be a uniformly convex real Banach space. For arbitrary $r>0$, let $B_r(0):=\{x\in E:\parallel x\parallel \le r\}$. Then, for any given sequence ${\{x_n\}}_{n=1}^{\mathrm{\infty }}\subset B_r(0)$ and for any given sequence ${\{{\lambda }_n\}}_{n=1}^{\mathrm{\infty }}$ of positive numbers such that ${\sum }_{i=1}^{\mathrm{\infty }}{\lambda }_i=1$, there exists a continuous strictly increasing convex function $g:[0,2r]\to \mathbb{R}$, $g(0)=0$ such that for any positive integers i, j with $i<j$, the following inequality holds

$${\parallel \sum _{n=1}^{\mathrm{\infty }}{\lambda }_n{x}_n\parallel }^2\le \sum _{n=1}^{\mathrm{\infty }}{\lambda }_n{\parallel x_n\parallel }^2-{\lambda }_i{\lambda }_{j}g(\parallel x_i-x_j\parallel ).$$

Lemma 2.11 [17]

Let E be a Banach space and $y\in E$. Let $f:E\to \mathbb{R}\cup \{+\mathrm{\infty }\}$ be a proper, convex and lower semi-continuous mapping with convex domain $D(f)$. If $\{x_n\}$ is a sequence in $D(f)$ such that $x_n\rightharpoonup x\in int(D(f))$ and ${lim}_{n\to \mathrm{\infty }}G(x_n,Jy)=G(x,Jy)$, then ${lim}_{n\to \mathrm{\infty }}\parallel x_n\parallel =\parallel x\parallel$.

For solving the equilibrium problem for a bifunction $F:C\times C\to \mathbb{R}$, let us assume that F satisfies the following conditions:

• (A1) $F(x,x)=0$ for all $x\in C$;

• (A2) F is monotone, i.e., $F(x,y)+F(y,x)\le 0$ for all $x,y\in C$;

• (A3) for each $x,y,z\in C$, ${lim}_{t\to 0}F(tz+(1-t)x,y)\le F(x,y)$;

• (A4) for each $x\in C$, $y\mapsto F(x,y)$ is convex and lower semicontinuous.

Lemma 2.12 [20]

Let C be a nonempty closed convex subset of a smooth, strictly convex and reflexive Banach space E, and let F be a bifunction of $C\times C$ into ℝ satisfying (A1)-(A4). Let $r>0$ and $x\in E$. Then, there exists $z\in C$ such that

$$F(z,y)+\frac{1}{r}〈y-z,Jz-Jx〉\ge 0,\mathrm{\forall }y\in C.$$

Lemma 2.13 [9, 21]

Let C be a nonempty closed convex subset of a smooth, strictly convex and reflexive Banach space E, and let F be a bifunction of $C\times C$ into ℝ satisfying (A1)-(A4). Let $r>0$ and $x\in E$. Define a mapping $T_r^F:E\to C$ as follows:

$$T_r^F(x)=\{z\in C:F(z,y)+\frac{1}{r}〈y-z,Jz-Jx〉\ge 0,\mathrm{\forall }y\in C\}$$

for all $z\in E$. Then, the following hold:

• 1. $T_r^F$ is single-valued;

• 2. $T_r^F$ is firmly nonexpansive mapping, i.e., for any $x,y\in E$,

  $$〈T_r^{F}x-T_r^{F}y,J{T}_r^{F}x-J{T}_r^{F}y〉\le 〈T_r^{F}x-T_r^{F}y,Jx-Jy〉;$$

• 3. $F(T_r^F)=EP(F)$;

• 4. $T_r^{F}x$ is relatively quasi-nonexpansive;

• 5. $EP(F)$ is closed and convex.

Lemma 2.14 [21]

Let C be a nonempty closed convex subset of a smooth, strictly convex and reflexive Banach space E, and let F be a bifunction of $C\times C$ into ℝ satisfying (A1)-(A4). Let $r>0$. Then for each $x\in E$ and $q\in F(T_r^F)$,

$$\varphi (q,T_r^{F}x)+\varphi (T_r^{F}x,x)\le \varphi (q,x).$$

An operator T in a Banach space E is said to be closed if $x_n\to x$ and $T{x}_n\to y$, then $Tx=y$.

3 Main result

Theorem 3.1 Let E be a uniformly smooth and strictly convex Banach space, which has the Kadec-Klee property, and let C be a nonempty closed convex subset of E. For each $k=1,2,\dots ,m$, let $F_k$ be a bifunction from $C\times C$ satisfying (A1)-(A4), and let ${\{T_i\}}_{i=0}^{\mathrm{\infty }}:C\to C$, $\mathrm{\forall }i\in \mathbb{N}$ be an infinite family of closed and asymptotically quasi-ϕ-nonexpansive mappings with sequence $\{k_{ni}\}\subset [1,\mathrm{\infty })$, $k_{ni}\to 1$ as $n\to \mathrm{\infty }$, where $T_0=I$. Assume that $T_i$, $\mathrm{\forall }i\in \mathbb{N}$ is asymptotically regular on C and $\mathrm{\Omega }=({\bigcap }_{i=0}^{\mathrm{\infty }}F(T_i))\cap ({\bigcap }_{k=1}^{m}EP(F_k))$ is nonempty and bounded. Let $f:E\to \mathbb{R}$ be a convex and lower semicontinuous mapping with $C\subset int(D(f))$, and suppose that ${\{x_n\}}_{n=0}^{\mathrm{\infty }}$ is a sequence generated by $x_0\in C$, $C_1=C$, $x_1={\mathrm{\Pi }}_{C_1}^f{x}_0$,

$$\{\begin{array}{c}{y}_n=J^{-1}\{{\sum }_{i=0}^{\mathrm{\infty }}{\alpha }_{ni}J{T}_i^n{x}_n\},\hfill \\ u_n=T_{r_{m,n}}^{F_m}{T}_{r_{m-1,n}}^{F_{m-1}}\cdots T_{r_{2,n}}^{F_2}{T}_{r_{1,n}}^{F_1}{y}_n,\hfill \\ C_{n+1}=\{z\in C_n:G(z,J{u}_n)\le G(z,J{x}_n)+(k_n-1)M_n\},\hfill \\ x_{n+1}={\mathrm{\Pi }}_{C_{n+1}}^f{x}_0,\hfill \end{array}$$

(3.1)

where J is the duality mapping on E, $M_n=sup\{\varphi (z,x_n):z\in \mathrm{\Omega }\}$ for each $n\ge 1$, $k_n={sup}_{i\ge 0}\{k_{ni}\}$, $\{{\alpha }_{ni}\}$ is a real sequence in [0,1] and ${\{r_{k,n}\}}_{n=1}^{\mathrm{\infty }}\subset (0,\mathrm{\infty })$, $k=1,2,\dots ,m$, satisfying the following conditions:

$$\begin{array}{r}\text{(a)}\sum _{i=0}^{\mathrm{\infty }}{\alpha }_{ni}=1,\mathrm{\forall }n\ge 1;\\ \text{(b)}\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}{\alpha }_{n0}{\alpha }_{ni}>0,\mathrm{\forall }i\in \mathbb{N};\\ \text{(c)}\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}{r}_{k,n}>0.\end{array}$$

Then the sequence $\{x_n\}$ converges strongly to ${\mathrm{\Pi }}_{\mathrm{\Omega }}^f{x}_0$.

Proof Step 1. We first show that $C_n$, $\mathrm{\forall }n\ge 1$ is nonempty, closed and convex.

Now, we show that $C_n$, $\mathrm{\forall }n\ge 1$ is closed and convex. It is obvious that $C_1=C$ is closed and convex. Suppose that $C_n$ is closed convex for some $n>1$. From the definition of $C_{n+1}$, we have $z\in C_{n+1}$, which implies that $G(z,J{u}_n)\le G(z,J{x}_n)+(k_n-1)M_n$. This is equivalent to

$$2(〈z,J{x}_n〉-〈z,J{u}_n〉)\le {\parallel x_n\parallel }^2-{\parallel u_n\parallel }^2+(k_n-1)M_n.$$

This implies that $C_{n+1}$ is closed convex for the same $n>1$. Hence, $C_n$ is closed and convex $\mathrm{\forall }n\ge 1$.

By taking ${\theta }_n^k=T_{r_{k,n}}^{F_k}{T}_{r_{k-1,n}}^{F_{k-1}}\cdots T_{r_{2,n}}^{F_2}{T}_{r_{1,n}}^{F_1}$, $k=1,2,\dots ,m$ and ${\theta }_n^0=I$ for all $n\ge 1$, we obtain $u_n={\theta }_n^m{y}_n$.

We next show that $\mathrm{\Omega }\subset C_n$, $\mathrm{\forall }n\ge 1$. From Lemma 2.13, we have that $T_{r_{k,n}}^{F_k}$, $k=1,2,\dots ,m$ is relatively nonexpansive mapping. For $n=1$, we have $\mathrm{\Omega }\subset C_1=C$. Now, assume that $\mathrm{\Omega }\subset C_n$ for some $n>1$. For each $x^{\ast }\in \mathrm{\Omega }$, we obtain

$$\begin{array}{rl}G(x^{\ast },J{u}_n)& =G(x^{\ast },J{\theta }_n^m{y}_n)\le G(x^{\ast },J{y}_n)\\ ={\parallel x^{\ast }\parallel }^2-2〈x^{\ast },J{y}_n〉+{\parallel y_n\parallel }^2+2\rho f\left(x^{\ast }\right)\\ \le {\parallel x^{\ast }\parallel }^2-2\sum _{i=0}^{\mathrm{\infty }}{\alpha }_{ni}〈x^{\ast },J{T}_i^n{x}_n〉+\sum _{i=0}^{\mathrm{\infty }}{\alpha }_{ni}{\parallel T_i^n{x}_n\parallel }^2+2\rho f\left(x^{\ast }\right)\\ =\sum _{i=0}^{\mathrm{\infty }}{\alpha }_{ni}\varphi (x^{\ast },T_i^n{x}_n)+2\rho f\left(x^{\ast }\right)\\ \le \sum _{i=0}^{\mathrm{\infty }}{\alpha }_{ni}{k}_{ni}\varphi (x^{\ast },x_n)+2\rho f\left(x^{\ast }\right)\\ =\sum _{i=0}^{\mathrm{\infty }}{\alpha }_{ni}(1+(k_{ni}-1))\varphi (x^{\ast },x_n)+2\rho f\left(x^{\ast }\right)\\ =G(x^{\ast },J{x}_n)+\sum _{i=0}^{\mathrm{\infty }}{\alpha }_{ni}(k_{ni}-1)\varphi (x^{\ast },x_n)\\ \le G(x^{\ast },J{x}_n)+(k_n-1)M_n.\end{array}$$

(3.2)

So, $x^{\ast }\in C_{n+1}$. It implies that $\mathrm{\Omega }\subset C_n$, $\mathrm{\forall }n\ge 1$, and the sequence ${\{x_n\}}_{n=0}^{\mathrm{\infty }}$ generated by (3.1) is well defined.

Step 2. We show that ${lim}_{n\to \mathrm{\infty }}G(x_n,J{x}_0)$ exists.

Since $f:E\to \mathbb{R}$ is a convex and lower semi-continuous, applying Lemma 2.7, we see that there exist $u^{\ast }\in E^{\ast }$ and $\alpha \in \mathbb{R}$ such that

$$f(y)\ge 〈y,u^{\ast }〉+\alpha ,\mathrm{\forall }y\in E.$$

It follows that

$$\begin{array}{rl}G(x_n,J{x}_0)& ={\parallel x_n\parallel }^2-2〈x_n,J{x}_0〉+{\parallel x_0\parallel }^2+2\rho f(x_n)\\ \ge {\parallel x_n\parallel }^2-2〈x_n,J{x}_0〉+{\parallel x_0\parallel }^2+2\rho 〈x_n,u^{\ast }〉+2\rho \alpha \\ ={\parallel x_n\parallel }^2-2〈x_n,J{x}_0-\rho u^{\ast }〉+{\parallel x_0\parallel }^2+2\rho \alpha \\ \ge {\parallel x_n\parallel }^2-2\parallel x_n\parallel \parallel J{x}_0-\rho u^{\ast }\parallel +{\parallel x_0\parallel }^2+2\rho \alpha \\ ={(\parallel x_n\parallel -\parallel J{x}_0-\rho u^{\ast }\parallel )}^2+{\parallel x_0\parallel }^2-{\parallel J{x}_0-\rho u^{\ast }\parallel }^2+2\rho \alpha .\end{array}$$

(3.3)

Since $x_n={\mathrm{\Pi }}_{C_n}^f{x}_0$, it follows from (3.3) that

$$G(x^{\ast },J{x}_0)\ge G(x_n,J{x}_0)\ge {(\parallel x_n\parallel -\parallel J{x}_0-\rho u^{\ast }\parallel )}^2+{\parallel x_0\parallel }^2-{\parallel J{x}_0-\rho u^{\ast }\parallel }^2+2\rho \alpha$$

for each $x^{\ast }\in F(T)$. This implies that ${\{x_n\}}_{n=0}^{\mathrm{\infty }}$ is bounded and so is ${\{G(x_n,J{x}_0)\}}_{n=0}^{\mathrm{\infty }}$. By the construction of $C_n$, we have that $C_{n+1}\subset C_n$ and $x_{n+1}={\mathrm{\Pi }}_{C_{n+1}}^f{x}_0\in C_n$. It follows from Lemma 2.8 that

$$\varphi (x_{n+1},x_n)+G(x_n,J{x}_0)\le G(x_{n+1},J{x}_0).$$

(3.4)

It is obvious that

$$\varphi (x_{n+1},x_n)\ge {(\parallel x_{n+1}\parallel -\parallel x_n\parallel )}^2\ge 0,$$

and so, ${\{G(x_n,J{x}_0)\}}_{n=0}^{\mathrm{\infty }}$ is nondecreasing. It follows that the limit of ${\{G(x_n,J{x}_0)\}}_{n=0}^{\mathrm{\infty }}$ exists.

Step 3. We prove that ${lim}_{n\to \mathrm{\infty }}\parallel J{x}_n-J{T}_j^n{x}_n\parallel =0$, $\mathrm{\forall }j\in \mathbb{N}$.

Now, since ${\{x_n\}}_{n=0}^{\mathrm{\infty }}$ is bounded in C, and E is reflexive, we may assume that $x_n\rightharpoonup p$, and since $C_n$ is closed and convex for each $n\ge 1$, it is easy to see that $p\in C_n$ for each $n\ge 1$. Again, since $x_n={\mathrm{\Pi }}_{C_n}^f{x}_0$, we obtain

$$G(x_n,J{x}_0)\le G(p,J{x}_0),\mathrm{\forall }n\ge 0.$$

Since

$$\begin{array}{rl}\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}G(x_n,J{x}_0)& =\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\{{\parallel x_n\parallel }^2-2〈x_n,J{x}_0〉+{\parallel x_0\parallel }^2+2\rho f(x_n)\}\\ \ge {\parallel p\parallel }^2-2〈p,J{x}_0〉+{\parallel x_0\parallel }^2+2\rho f(p)=G(p,J{x}_0).\end{array}$$

Then, we obtain

$$G(p,J{x}_0)\le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}G(x_n,J{x}_0)\le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}G(x_n,J{x}_0)\le G(p,J{x}_0).$$

This implies that

$$\underset{n\to \mathrm{\infty }}{lim}G(x_n,J{x}_0)=G(p,J{x}_0).$$

By Lemma 2.11, we obtain that ${lim}_{n\to \mathrm{\infty }}\parallel x_n\parallel =\parallel p\parallel$. In view of Kadec-Klee property of E, we have that ${lim}_{n\to \mathrm{\infty }}{x}_n=p$.

By the construction of $C_n$, we have that $C_{n+1}\subset C_n$ and $x_{n+1}={\mathrm{\Pi }}_{C_{n+1}}^f{x}_0\in C_{n+1}$. It follows that

$$\varphi (x_{n+1},u_n)\le \varphi (x_{n+1},x_n)+(k_n-1)M_n.$$

Now, (3.4) implies that

$$\varphi (x_{n+1},u_n)\le \varphi (x_{n+1},x_n)+(k_n-1)M_n\le G(x_{n+1},J{x}_0)-G(x_n,J{x}_0)+(k_n-1)M_n.$$

(3.5)

Taking the limit as $n\to \mathrm{\infty }$ in (3.5), we obtain

$$\underset{n\to \mathrm{\infty }}{lim}\varphi (x_{n+1},x_n)=0.$$

Therefore,

$$\underset{n\to \mathrm{\infty }}{lim}\varphi (x_{n+1},u_n)=0.$$

It then yields that ${lim}_{n\to \mathrm{\infty }}(\parallel x_{n+1}\parallel -\parallel u_n\parallel )=0$. Since ${lim}_{n\to \mathrm{\infty }}\parallel x_{n+1}\parallel =\parallel p\parallel$, we have

$$\underset{n\to \mathrm{\infty }}{lim}\parallel u_n\parallel =\parallel p\parallel .$$

(3.6)

Hence,

$$\underset{n\to \mathrm{\infty }}{lim}\parallel J{u}_n\parallel =\parallel Jp\parallel .$$

This implies that ${\{\parallel J{u}_n\parallel \}}_{n=0}^{\mathrm{\infty }}$ is bounded in $E^{\ast }$. Since E is reflexive, and so $E^{\ast }$ is reflexive, we can then assume that $J{u}_n\rightharpoonup f_0\in E^{\ast }$. In view of reflexivity of E, we see that $J(E)=E^{\ast }$. Hence, there exists $x\in E$ such that $Jx=f_0$. Since

$$\varphi (x_{n+1},u_n)={\parallel x_{n+1}\parallel }^2-2〈x_{n+1},J{u}_n〉+{\parallel u_n\parallel }^2={\parallel x_{n+1}\parallel }^2-2〈x_{n+1},J{u}_n〉+{\parallel J{u}_n\parallel }^2.$$

Taking ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}$ for both sides of the equality above, yields that

$$0\ge {\parallel p\parallel }^2-2〈p,f_0〉+{\parallel f_0\parallel }^2={\parallel p\parallel }^2-2〈p,Jx〉+{\parallel Jx\parallel }^2={\parallel p\parallel }^2-2〈p,Jx〉+{\parallel x\parallel }^2=\varphi (p,x).$$

That is, $p=x$. This implies that $f_0=Jp$, and so, $J{u}_n\rightharpoonup Jp$. It follows from ${lim}_{n\to \mathrm{\infty }}\parallel J{u}_n\parallel =\parallel Jp\parallel$ and Kadec-Klee property of $E^{\ast }$ (this is because $E^{\ast }$ is uniformly convex) that

$$J{u}_n\to Jp.$$

Note that $J^{-1}:E^{\ast }\to E$ is hemi-continuous (this is because E is a uniformly smooth and strictly convex Banach space with a strictly convex dual $E^{\ast }$), it follows that $u_n\rightharpoonup p$. Since (3.6) and E have the Kadec-Klee property, we obtain that

$$\underset{n\to \mathrm{\infty }}{lim}{u}_n=p.$$

(3.7)

It follows that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_n-u_n\parallel =0.$$

(3.8)

Since J is uniformly norm-to-norm continuous on any bounded sets, we have

$$\underset{n\to \mathrm{\infty }}{lim}\parallel J{x}_n-J{u}_n\parallel =0.$$

(3.9)

Let $r:={sup}_{n,i\ge 0}\{\parallel T_i^n{x}_n\parallel \}$. Since E is uniformly smooth, we know that $E^{\ast }$ is uniformly convex. Then from Lemma 2.10, we have

$$\begin{array}{rl}G(x^{\ast },J{u}_n)=& G(x^{\ast },J{\theta }_n^m{y}_n)\\ \le & G(x^{\ast },J{y}_n)\\ =& {\parallel x^{\ast }\parallel }^2-2〈x^{\ast },\sum _{i=0}^{\mathrm{\infty }}{\alpha }_{ni}J{T}_i^n{x}_n〉+{\parallel \sum _{i=0}^{\mathrm{\infty }}{\alpha }_{ni}J{T}_i^n{x}_n\parallel }^2+2\rho f\left(x^{\ast }\right)\\ \le & {\parallel x^{\ast }\parallel }^2-2\sum _{i=0}^{\mathrm{\infty }}{\alpha }_{ni}〈x^{\ast },J{T}_i^n{x}_n〉+\sum _{i=0}^{\mathrm{\infty }}{\alpha }_{ni}{\parallel J{T}_i^n{x}_n\parallel }^2\\ -{\alpha }_{nk}{\alpha }_{nj}g\left(\parallel J{T}_k^n{x}_n-J{T}_j^n{x}_n\parallel \right)+2\rho f\left(x^{\ast }\right)\\ =& \sum _{i=0}^{\mathrm{\infty }}{\alpha }_{ni}\varphi (x^{\ast },T_i^n{x}_n)+2\rho f\left(x^{\ast }\right)-{\alpha }_{nk}{\alpha }_{nj}g\left(\parallel J{T}_k^n{x}_n-J{T}_j^n{x}_n\parallel \right)\\ \le & \sum _{i=0}^{\mathrm{\infty }}{\alpha }_{ni}{k}_{ni}\varphi (x^{\ast },x_n)+2\rho f\left(x^{\ast }\right)-{\alpha }_{nk}{\alpha }_{nj}g\left(\parallel J{T}_k^n{x}_n-J{T}_j^n{x}_n\parallel \right)\\ \le & G(x^{\ast },J{x}_n)+(k_n-1)M_n-{\alpha }_{nk}{\alpha }_{nj}g\left(\parallel J{T}_k^n{x}_n-J{T}_j^n{x}_n\parallel \right).\end{array}$$

(3.10)

Taking $k=0$ and for any j in (3.10), we have

$${\alpha }_{n0}{\alpha }_{nj}g\left(\parallel J{x}_n-J{T}_j^n{x}_n\parallel \right)\le G(x^{\ast },J{x}_n)-G(x^{\ast },J{u}_n)+(k_n-1)M_n\to 0.$$

It follows from the property of g that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel J{x}_n-J{T}_j^n{x}_n\parallel =0.$$

(3.11)

Step 4. Now we prove that $p\in \mathrm{\Omega }$.

(a) First, we prove that $p\in {\bigcap }_{i=0}^{\mathrm{\infty }}F(T_i)$.

Since $x_n\to p$ and J is uniformly norm-to-norm continuous on bounded sets, we see that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel J{x}_n-Jp\parallel =0.$$

(3.12)

We observe from (3.11) and (3.12) that

$$\parallel J{T}_j^n{x}_n-Jp\parallel \le \parallel J{x}_n-J{T}_j^n{x}_n\parallel +\parallel J{x}_n-Jp\parallel \to 0,n\to \mathrm{\infty }.$$

Since $J^{-1}$ is hemi-continuous, it follows that $T_j^n{x}_n\rightharpoonup p$. On the other hand, since

$$|\parallel T_j^n{x}_n\parallel -\parallel p\parallel |=|\parallel J{T}_j^n{x}_n\parallel -\parallel Jp\parallel |\le \parallel J{T}_j^n{x}_n-Jp\parallel ,$$

and this implies that $\parallel T_j^n{x}_n\parallel \to \parallel p\parallel$ as $n\to \mathrm{\infty }$. Since E enjoys the Kadec-Klee property, we obtain that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel T_j^n{x}_n-p\parallel =0.$$

Note that

$$\parallel T_j^{n+1}{x}_n-p\parallel \le \parallel T_j^{n+1}{x}_n-T_j^n{x}_n\parallel +\parallel T_j^n{x}_n-p\parallel .$$

(3.13)

It follows from the asymptotic regularity of T and (3.13) that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel T_j^{n+1}{x}_n-p\parallel =0.$$

That is, $T_j{T}_j^n{x}_n-p\to 0$ as $n\to \mathrm{\infty }$. It follows from the closeness of $T_j$ that $T_{j}p=p$, $\mathrm{\forall }j\in \mathbb{N}$, i.e., $p\in {\bigcap }_{i=0}^{\mathrm{\infty }}F(T_i)$.

(b) Next, we prove that $p\in {\bigcap }_{k=1}^{m}EP(F_k)$.

From (3.2), we obtain

$$\begin{array}{rl}\varphi (x^{\ast },u_n)& =\varphi (x^{\ast },{\theta }_n^m{y}_n)=\varphi (x^{\ast },T_{r_{m,n}}^{F_m}{\theta }_n^{m-1}{y}_n)\\ \le \varphi (x^{\ast },{\theta }_n^{m-1}{y}_n)\le \cdots \le \varphi (x^{\ast },y_n)\\ \le \varphi (x^{\ast },x_n)+(k_n-1)M_n.\end{array}$$

Next, we show that ${\theta }_n^k{y}_n\to p$ as $n\to \mathrm{\infty }$, for each $k\in \{0,1,\dots ,m\}$.

We have proved that $k=m$, ${\theta }_n^k{y}_n=u_n\to p$.

Suppose that ${\theta }_n^k{y}_n\to p$ as $n\to \mathrm{\infty }$ for some k. Since $x^{\ast }\in {\bigcap }_{k=1}^{m}EP(F_k)={\bigcap }_{k=1}^{m}F(T_{r_{k,n}}^{F_k})$ for all $n\ge 1$, it follows from Lemma 2.14 that

$$\begin{array}{rl}\varphi ({\theta }_n^k{y}_n,{\theta }_n^{k-1}{y}_n)& =\varphi (T_{r_{k,n}}^{F_k}{\theta }_n^{k-1}{y}_n,{\theta }_n^{k-1}{y}_n)\\ \le \varphi (x^{\ast },{\theta }_n^{k-1}{y}_n)-\varphi (x^{\ast },{\theta }_n^k{y}_n)\\ \le \varphi (x^{\ast },x_n)-\varphi (x^{\ast },{\theta }_n^k{y}_n)+(k_n-1)M_n.\end{array}$$

Hence, we have

$$\underset{n\to \mathrm{\infty }}{lim}\varphi ({\theta }_n^k{y}_n,{\theta }_n^{k-1}{y}_n)=0.$$

From (2.5), we see that $\parallel {\theta }_n^k{y}_n\parallel -\parallel {\theta }_n^{k-1}{y}_n\parallel \to 0$ as $n\to \mathrm{\infty }$. From assumption, we have ${\theta }_n^k{y}_n\to p$ as $n\to \mathrm{\infty }$, so

$$\parallel {\theta }_n^{k-1}{y}_n\parallel \to \parallel p\parallel \text{as }n\to \mathrm{\infty }.$$

It follows that

$$\parallel J{\theta }_n^{k-1}{y}_n\parallel \to \parallel Jp\parallel \text{as }n\to \mathrm{\infty }.$$

(3.14)

This implies that ${\{\parallel J{\theta }_n^{k-1}{y}_n\parallel \}}_{n=0}^{\mathrm{\infty }}$ is bounded in $E^{\ast }$. Since E is reflexive, and so $E^{\ast }$ is reflexive, we can then assume that $J{\theta }_n^{k-1}{y}_n\rightharpoonup f_{k-1}\in E^{\ast }$. In view of reflexivity of E, we see that $J(E)=E^{\ast }$. Hence, there exists $x^{k-1}\in E$ such that $J{x}^{k-1}=f_{k-1}$. Since

$$\begin{array}{rl}\varphi ({\theta }_n^k{y}_n,{\theta }_n^{k-1}{y}_n)& ={\parallel {\theta }_n^k{y}_n\parallel }^2-2〈{\theta }_n^k{y}_n,J{\theta }_n^{k-1}{y}_n〉+{\parallel {\theta }_n^{k-1}{y}_n\parallel }^2\\ ={\parallel {\theta }_n^k{y}_n\parallel }^2-2〈{\theta }_n^k{y}_n,J{\theta }_n^{k-1}{y}_n〉+{\parallel J{\theta }_n^{k-1}{y}_n\parallel }^2.\end{array}$$

Taking ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}$ for both sides of the equality above, yields that

$$\begin{array}{rl}0& \ge {\parallel p\parallel }^2-2〈p,f_{k-1}〉+{\parallel f_{k-1}\parallel }^2\\ ={\parallel p\parallel }^2-2〈p,J{x}^{k-1}〉+{\parallel J{x}^{k-1}\parallel }^2\\ ={\parallel p\parallel }^2-2〈p,J{x}^{k-1}〉+{\parallel x^{k-1}\parallel }^2\\ =\varphi (p,x^{k-1}).\end{array}$$

That is, $p=x^{k-1}$. This implies that $f_{k-1}=Jp$ and so $J{\theta }_n^{k-1}{y}_n\rightharpoonup Jp$. It follows from ${lim}_{n\to \mathrm{\infty }}\parallel J{\theta }_n^{k-1}{y}_n\parallel =\parallel Jp\parallel$ and Kadec-Klee property of $E^{\ast }$ (this is because $E^{\ast }$ is uniformly convex) that

$$J{\theta }_n^{k-1}{y}_n\to Jp.$$

Note that $J^{-1}:E^{\ast }\to E$ is hemi-continuous (this is because E is a uniformly smooth and strictly convex Banach space with a strictly convex dual $E^{\ast }$), it follows that ${\theta }_n^{k-1}{y}_n\rightharpoonup p$. Since (3.14) and E have the Kadec-Klee property, we obtain that

$$\underset{n\to \mathrm{\infty }}{lim}{\theta }_n^{k-1}{y}_n=p.$$

Hence, ${lim}_{n\to \mathrm{\infty }}{\theta }_n^k{y}_n=p$ and ${lim}_{n\to \mathrm{\infty }}J{\theta }_n^k{y}_n=Jp$, for each $k\in \{0,1,\dots ,m\}$. That is,

$$\underset{n\to \mathrm{\infty }}{lim}\parallel {\theta }_n^k{y}_n-{\theta }_n^{k-1}{y}_n\parallel =0,k=1,2,\dots ,m$$

and

$$\underset{n\to \mathrm{\infty }}{lim}\parallel J{\theta }_n^k{y}_n-J{\theta }_n^{k-1}{y}_n\parallel =0,k=1,2,\dots ,m.$$

Since ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{r}_{k,n}>0$, $k=1,2,\dots ,m$,

$$\underset{n\to \mathrm{\infty }}{lim}\frac{\parallel J{\theta }_n^k{y}_n-J{\theta }_n^{k-1}{y}_n\parallel }{r_{k,n}}=0.$$

(3.15)

By Lemma 2.13, we have that for each $k=1,2,\dots ,m$,

$$F_k({\theta }_n^k{y}_n,y)+\frac{1}{r_{k,n}}〈y-{\theta }_n^k{y}_n,J{\theta }_n^k{y}_n-J{\theta }_n^{k-1}{y}_n〉\ge 0,\mathrm{\forall }y\in C.$$

Furthermore, using (A2), we obtain

$$\frac{1}{r_{k,n}}〈y-{\theta }_n^k{y}_n,J{\theta }_n^k{y}_n-J{\theta }_n^{k-1}{y}_n〉\ge F_k(y,{\theta }_n^k{y}_n).$$

By (A4), (3.15) and ${\theta }_n^k{y}_n\to p$, we have for each $k=1,2,\dots ,m$,

$$F_k(y,p)\le 0,\mathrm{\forall }y\in C.$$

For fixed $y\in C$, let $z_t=ty+(1-t)p$ for all $t\in (0,1]$. This implies that $z_t\in C$. This yields that $F_k(z_t,p)\le 0$. It follows from (A1) and (A4) that

$$0=F_k(z_t,z_t)\le t{F}_k(z_t,y)+(1-t)F_k(z_t,p)\le t{F}_k(z_t,y),$$

and hence

$$0\le F_k(z_t,y).$$

From condition (A3), we obtain

$$F_k(p,y)\ge 0,\mathrm{\forall }y\in C.$$

This implies that $p\in EP(F_k)$, $k=1,2,\dots ,m$. Thus, $p\in {\bigcap }_{k=1}^{m}EP(F_m)$.

Hence, we have $p\in \mathrm{\Omega }=({\bigcap }_{k=1}^{m}EP(F_m))\cap ({\bigcap }_{i=0}^{\mathrm{\infty }}F(T_i))$.

Step 5. Finally, we prove that $p={\mathrm{\Pi }}_{\mathrm{\Omega }}^f{x}_0$.

Since $\mathrm{\Omega }=({\bigcap }_{k=1}^{m}EP(F_m))\cap ({\bigcap }_{i=0}^{\mathrm{\infty }}F(T_i))$ is a closed and convex set, from Lemma 2.6, we know that ${\mathrm{\Pi }}_{\mathrm{\Omega }}^f{x}_0$ is single-valued and denoted $\omega ={\mathrm{\Pi }}_{\mathrm{\Omega }}^f{x}_0$. Since $x_n={\mathrm{\Pi }}_{C_n}^f{x}_0$ and $\omega \in \mathrm{\Omega }\subset C_n$, we have

$$G(x_n,J{x}_0)\le G(\omega ,J{x}_0),\mathrm{\forall }n\ge 0.$$

We know that $G(\xi ,\varphi )$ is convex and lower semi-continuous with respect to ξ when ϕ is fixed. This implies that

$$G(p,J{x}_0)\le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}G(x_n,J{x}_0)\le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}G(x_n,J{x}_0)\le G(\omega ,J{x}_0).$$

From the definition of ${\mathrm{\Pi }}_{\mathrm{\Omega }}^f{x}_0$ and $p\in \mathrm{\Omega }$, we see that $p=\omega$. This completes the proof. □

Corollary 3.2 Let E be a uniformly smooth and strictly convex Banach space, which has the Kadec-Klee property, and let C be a nonempty closed convex subset of E. For each $k=1,2,\dots ,m$, let $F_k$ be a bifunction from $C\times C$ satisfying (A1)-(A4), and let ${\{T_i\}}_{i=0}^{\mathrm{\infty }}:C\to C$, $\mathrm{\forall }i\in \mathbb{N}$ be an infinite family of closed and asymptotically quasi-ϕ-nonexpansive mappings with sequence $\{k_{ni}\}\subset [1,\mathrm{\infty })$, $k_{ni}\to 1$ as $n\to \mathrm{\infty }$, where $T_0=I$. Assume that $T_i$, $\mathrm{\forall }i\in \mathbb{N}$ is asymptotically regular on C, and $\mathrm{\Omega }=({\bigcap }_{i=0}^{\mathrm{\infty }}F(T_i))\cap ({\bigcap }_{k=1}^{m}EP(F_k))$ is nonempty and bounded. Suppose that ${\{x_n\}}_{n=0}^{\mathrm{\infty }}$ is generated by $x_0\in C$, $C_1=C$, $x_1={\mathrm{\Pi }}_{C_1}^f{x}_0$,

$$\{\begin{array}{c}{y}_n=J^{-1}\{{\sum }_{i=0}^{\mathrm{\infty }}{\alpha }_{ni}J{T}_i^{n}x\},\hfill \\ u_n=T_{r_{m,n}}^{F_m}{T}_{r_{m-1,n}}^{F_{m-1}}\cdots T_{r_{2,n}}^{F_2}{T}_{r_{1,n}}^{F_1}{y}_n,\hfill \\ C_{n+1}=\{z\in C_n:\varphi (z,J{u}_n)\le \varphi (z,J{x}_n)+(k_n-1)M_n\},\hfill \\ x_{n+1}={\mathrm{\Pi }}_{C_{n+1}}{x}_0,\hfill \end{array}$$

where J is the duality mapping on E, $M_n=sup\{\varphi (z,x_n):z\in \mathrm{\Omega }\}$ for each $n\ge 1$, $k_n={sup}_{i\ge 0}\{k_{ni}\}$, $\{{\alpha }_{ni}\}$ is a real sequence in $[0,1]$ and ${\{r_{k,n}\}}_{n=1}^{\mathrm{\infty }}\subset (0,\mathrm{\infty })$, $k=1,2,\dots ,m$, satisfying the following conditions:

$$\begin{array}{r}\text{(a)}\sum _{i=0}^{\mathrm{\infty }}{\alpha }_{ni}=1,\mathrm{\forall }n\ge 1;\\ \text{(b)}\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}{\alpha }_{n0}{\alpha }_{ni}>0,\mathrm{\forall }i\in \mathbb{N};\\ \text{(c)}\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}{r}_{k,n}>0.\end{array}$$

Then the sequence $\{x_n\}$ converges strongly to ${\mathrm{\Pi }}_{\mathrm{\Omega }}{x}_0$.